The First Fundamental Theorem in Finite-State Markets

From No Arbitrage to a Positive Pricing Rule

In a finite market, every portfolio’s terminal gain is a vector: one component for each possible terminal state. Arbitrage means that this vector is nonnegative in every state and positive in at least one. A strictly positive pricing rule assigns positive weight to every state. If it assigns zero value to every attainable zero-cost gain, an arbitrage is impossible: a nonnegative, nonzero gain would have strictly positive weighted value. The substantive direction is to prove that no arbitrage guarantees such a positive rule.

We work on a finite filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_t)_{t=0}^{T},\mathbb{P})\), where \(\Omega=\{\omega_1,\ldots,\omega_m\}\), \(\mathcal{F}_T=2^\Omega\), and \(\mathbb{P}(\{\omega_i\})>0\) for every \(i\). Here \(T\) is a positive integer. This is a discrete-time market, so no Brownian motion or stochastic integral is used. There are \(d\) traded assets with adapted prices \(S_t=(S_t^1,\ldots,S_t^d)\), and a strictly positive, deterministic bank account \(A_t\), with \(A_0=1\). Define discounted asset prices by \(\widetilde S_t=S_t/A_t\), componentwise.

A strategy \(\theta=(\theta_1,\ldots,\theta_T)\) is predictable: \(\theta_t\in\mathbb{R}^d\) is \(\mathcal{F}_{t-1}\)-measurable and records the holdings over period \((t-1,t]\). Its discounted gain from zero initial cost is the terminal random variable

In this frictionless model, such gains are the discounted gains of self-financing trading strategies. Because the state space and horizon are finite, every strategy and gain is finite-valued and integrable. The set \(G\subseteq\mathbb{R}^m\) of all gain vectors is a linear subspace: adding or multiplying strategies adds or multiplies their gains. A vector inequality such as \(g\geq0\) means \(g_i\geq0\) in every state.

Equivalence means that \(\mathbb{Q}(\{\omega_i\})>0\) for every state. In this finite model, the martingale condition means that \(\mathbb{E}^{\mathbb{Q}}[\widetilde S_t\mid\mathcal{F}_{t-1}]=\widetilde S_{t-1}\) for each \(t\), component by component. We will prove the theorem by first establishing the finite-dimensional separation fact that supplies strictly positive state weights.

The Separation Lemma

Proof. First suppose \(q\in L^\perp\) and \(q_i>0\) for every \(i\). If \(x\in L\cap\mathbb{R}_+^m\), then \(q\cdot x=0\) because \(x\in L\). But if \(x\neq0\), at least one coordinate is positive, all are nonnegative, and all coordinates of \(q\) are positive; hence \(q\cdot x=\sum_iq_ix_i>0\), a contradiction. Thus the intersection contains only \(0\).

For the converse, assume \(L\cap\mathbb{R}_+^m=\{0\}\). Let \(C=\{x\in\mathbb{R}_+^m:\sum_i x_i=1\}\). This set is nonempty and compact. It is disjoint from \(L\): any point in \(C\cap L\) would be a nonzero point in \(L\cap\mathbb{R}_+^m\), since its coordinates sum to \(1\). Let \(P\) be the orthogonal projection onto \(L^\perp\). The continuous function \(x\mapsto\lVert Px\rVert^2\) attains a minimum on \(C\), at some \(c\in C\). This minimum is positive: \(Pc=0\) would imply \(c\in L\), contrary to \(C\cap L=\varnothing\). Set \(q=Pc\), so \(q\in L^\perp\) and \(\lVert q\rVert^2>0\).

For any \(x\in C\) and \(0\leq\varepsilon\leq1\), the point \(c+\varepsilon(x-c)\) belongs to \(C\), by convexity. Minimality at \(c\) gives \(\lVert P(c+\varepsilon(x-c))\rVert^2\geq\lVert Pc\rVert^2\). Expanding the left side, dividing by \(\varepsilon>0\), and letting \(\varepsilon\) decrease to zero yields \(2q\cdot P(x-c)\geq0\). Since \(q\in L^\perp\), \(q\cdot Py=q\cdot y\) for every \(y\): the difference \(y-Py\) lies in \(L\). Therefore \(q\cdot x\geq q\cdot c=\lVert q\rVert^2>0\). Each coordinate vector \(e_i\) belongs to \(C\), so \(q_i=q\cdot e_i\geq\lVert q\rVert^2>0\). This constructs the required strictly positive vector and proves the converse. \(\square\)

The First Fundamental Theorem

Proof. Suppose first that an equivalent martingale measure \(\mathbb{Q}\) exists. For every predictable strategy, the martingale property gives \(\mathbb{E}^{\mathbb{Q}}[G(\theta)]=0\). Indeed, at each time \(t\), \(\theta_t\) is measurable with respect to \(\mathcal{F}_{t-1}\), so conditioning on that sigma-algebra gives \(\mathbb{E}^{\mathbb{Q}}[\theta_t\cdot(\widetilde S_t-\widetilde S_{t-1})]=0\); summing these finitely many equalities proves the claim. If an attainable gain were nonnegative in every state and positive in at least one, its expectation under \(\mathbb{Q}\) would be strictly positive, because \(\mathbb{Q}\) assigns positive probability to every state. This contradicts zero expectation. Thus there is no arbitrage.

Now suppose there is no arbitrage. The gain subspace \(G\) then satisfies \(G\cap\mathbb{R}_+^m=\{0\}\). By the separation lemma, there is a vector \(q\in G^\perp\) with \(q_i>0\) for every state. Define \(Z=\sum_{i=1}^m q_i>0\), and set \(\mathbb{Q}(\{\omega_i\})=q_i/Z\). These probabilities are positive and sum to one, so \(\mathbb{Q}\) is a probability measure equivalent to \(\mathbb{P}\). Since \(q\) annihilates every gain vector, \(\mathbb{E}^{\mathbb{Q}}[G(\theta)]=0\) for every predictable strategy.

It remains to check the conditional martingale property, rather than only an unconditional expectation. Fix a time \(t\), asset \(j\), and event \(D\in\mathcal{F}_{t-1}\). Choose a strategy that holds one unit of asset \(j\) during \((t-1,t]\) on \(D\), and holds zero in every other period and state. It is predictable, and its gain is \(\mathbf{1}_D(\widetilde S_t^j-\widetilde S_{t-1}^j)\). Its expectation under \(\mathbb{Q}\) is zero, so

On each atom \(D\) of \(\mathcal{F}_{t-1}\), this equality says that the conditional average of the increment is zero. Each nonempty atom has positive \(\mathbb{Q}\)-probability, since it contains at least one terminal state and every terminal state has positive \(\mathbb{Q}\)-probability. Thus \(\mathbb{E}^{\mathbb{Q}}[\widetilde S_t^j\mid\mathcal{F}_{t-1}]=\widetilde S_{t-1}^j\) on every atom. This holds for every asset and time, so the discounted prices are \(\mathbb{Q}\)-martingales. The measure is equivalent to \(\mathbb{P}\), completing the proof. \(\square\)

The proof identifies the key role of positivity. Orthogonality alone gives a linear pricing rule that values every attainable gain at zero. Strict positivity makes the rule assign positive value to every nonzero nonnegative payoff, which is what rules out arbitrage and ensures equivalence rather than merely absolute continuity.

Worked Examples

State Prices and a Common Pitfall

For any terminal payoff \(Y\), the state-price vector associated with an equivalent martingale measure is \(\pi_i=(A_0/A_T)\mathbb{Q}(\{\omega_i\})\). Its time-zero risk-neutral value is \(\sum_i\pi_iY(\omega_i)=A_0\mathbb{E}^{\mathbb{Q}}[Y/A_T]\). This formula prices any payoff by expectation; it does not assert that every payoff can be replicated by a trading strategy. The theorem concerns the existence of a positive pricing rule consistent with traded assets, not whether the market is complete.

Common mistake: accepting zero state weights. In the second example, \(q=0\) solves the martingale equation, but it does not give an equivalent martingale measure. Treating that zero as acceptable hides the arbitrage: the portfolio pays \(10\) in precisely the state assigned zero weight and \(0\) otherwise. Equivalence requires strictly positive probability in every state, not merely nonnegative weights summing to one.

The finite-state hypothesis matters to this proof because the attainable gains form a subspace of a finite-dimensional vector space, where the separation lemma applies as shown. The theorem proved here is for this finite, discrete-time setting. The corresponding questions involving admissibility and limiting notions of arbitrage in more general markets require additional hypotheses and arguments.